Question

2. Use the literature value of K, for acetic acid to calculate the percent ionization of a 0.40M solution.

Answer 1

The percent ionization is 0,66%.

Step-by-step explanation:

K for acetic acid = 1,74x10*-5

CH3COOH (ac) + H2O(l) ⇄ H3O+ (ac) + CH3COO- (ac)

initial. 0,40 M ---- ------

react. X X X

eq. (0,4 -X) X X

This is the expression for K:

Ka = ( [H3O+] . [CH3COO-] ) / [CH3COOH]

1,74x10*-5 = (X . X) / 0,40 - X

Since the X in the numerator is going to be a very small number, since my Ka is in the order of -5, I can despise it in order to avoid the quadratic equation. You can't always do that!.. Be careful and pay attention to the value of Ka.

`√(1,74x10*-5x0,40)\\` = X

X = 2,64x10*-3

% ionization is (X /acid concentration) x 100

(2,64x10*-3 / 0,4) x 100 = 0,66%